未定義函數eval（） - PHP

Question

我想用那樣的eval（）函數：

$foo = 'eval'; 
$bar = 'echo 1;'; 
$foo($bar); 

但我發現了一個錯誤：致命錯誤：調用未定義的函數eval（）

，由於下面的代碼工作它

$foo = 'base64_encode'; 
$bar = 'foobar'; 
echo $foo($bar); 

誰能幫助奇怪？

Answer 1

從eval documentation：

Note: Because this is a language construct and not a function, it cannot be called using variable functions .

繼在注意鏈接，你還會發現：

Variable functions won't work with language constructs such as echo , print , unset() , isset() , empty() , include , require and the like. Utilize wrapper functions to make use of any of these constructs as variable functions.